Answer
$\frac{\partial z}{\partial x}=\frac{-x}{3z}$
$\frac{\partial z}{\partial y}=\frac{-2y}{3z}$
Work Step by Step
Given $x^2+2y^2+3z^2=1$, w us impliit iffrntiation to fin $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$:
To find $\frac{\partial z}{\partial x}$, simply differentiate all terms with respect to $x$:
$\frac{\partial}{\partial x}(x^2+2y^2+3z^2)=2x+0+6z \frac{\partial z}{\partial x}=0$
The derivative of the L.H.S is equal to the derivative of the R.H.S. Given that the r.h.s was a constant, it's derivative is equal to zero. Thus:
$\frac{\partial z}{\partial x}=\frac{-2x}{6z}=\frac{-x}{3z}$
A similar procedure applies for the partial derivative with respect to $y$:
$\frac{\partial z}{\partial y}(x^2+2y^2+3z^2)=0+4y+6z \frac{\partial z}{\partial y}=0$
$\frac{\partial z}{\partial y}=\frac{-4y}{6z}=\frac{-2y}{3z}$
